Indian Statistical Institute, ISI BStat & BMath 2021 UGB Solutions & Discussions

ISI BMath & BSTAT 2021 Subjective Questions UGB: solutions and discussions

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Indian Statistical Institute,ISI BMATH/BSTAT 2021 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGA Objective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 All UGB Subjective Solutions: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Answer Keys: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGA Question Paper: Click Here

Indian Statistical Institute,ISI BMATH/BSTAT 2020 UGB Question Paper: Click Here

Problem 1.

There are three cities each of which has exactly the same number of citizens, say n. Every. citizen in each city has exactly a total of n+1 friends in the other two cities. Show that there exist three people, one from each city, such that they are friends. We assume that friendship is mutual (that is, a symmetric relation).

Topic: The Extremal Principle
Difficulty level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 1 UGB Solution

Solution:

From the 3n citizens, take one who has a maximum number k of acquaintances from one of the two other cities. Suppose it is citizen A from the first city, who knows k citizens from the second city. Then A knows (n+1-k) citizens from the third city, n+1-k \geq 1 since k \leq n. Consider citizen B from the third city, who knows A. If B knows at least one citizen C from the k acquaintances of A in the second city, then \{A, B, C\} is a triple of mutual acquaintances. But if B knows none of the k acquaintances of A in the second city, then, in this city he does not know more than (n-k) citizens, and hence, in the first city, he does not know less than n+1-(n-k)=k+1 citizens, which contradicts the choice of k.

Video Solution:

Problem 2.

Let f: \mathbb{Z} \rightarrow \mathbb{Z} be a function satisfying f(0) \neq 0=f(1). Assume also that f satisfies equations (A) and (B) below.

    \[\begin{aligned} f(x y) &=f(x)+f(y)-f(x) f(y) \\ f(x-y) f(x) f(y) &=f(0) f(x) f(y) \end{aligned}\]

for all integers x, y.
(i) Determine explicitly the set \{f(a): a \in \mathbb{Z}\}.
(ii) Assuming that there is a non-zero integer a such that f(a) \neq 0, prove that the set \{b: f(b) \neq 0\} is infinite.

Topic: Functional Equations
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 2 UGB Solution.

Solution:

Let x=y=0 in (A):

    \[f(0)=f(0)+f(0)-f(0)^2, f(0)^2=f(0).\]

Since f(0)\ne 0 we deduce f(0)=1.

Let y=0 in (B):

    \[f(x)^2f(0)=f(x)f(0)^2, f(x)^2=f(x).\]

Either f(x)=0 or f(x)=1.

The function f takes only two values 0 and 1 (both values are really in the range of f, because f(0)=1, f(1)=0).

\{f(a):a\in\mathbb{Z}\}=\{0,1\}.

Assume that there exist a non-zero a\in\mathbb{Z} such that f(a)\ne 0. Then a\ne 1. Take x=y=-1 in (A):

    \[0=f(1)=2f(-1)-f(-1)^2.\]

It follows that f(-1)\ne 1, hence f(-1)=0 and a\ne -1. So, |a|>1. Further, from (A) we find that if f(a)=f(b)=1, then

    \[f(ab)=f(a)+f(b)-f(a)f(b)=1.\]

By induction, f(a^k)=1 for all k\geq 1 and the set
\{b:f(b)\ne 0\} is infinite.

Video Solution:

Problem 3.

Prove that every positive rational number can be expressed uniquely as a finite sum of the form

    \[a_{1}+\frac{a_{2}}{2 !}+\frac{a_{3}}{3 !}+\cdots+\frac{a_{n}}{n !},\]

where a_{n} are integers such that 0 \leq a_{n} \leq n-1 for all n>1.

Topic: Number Theory
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 3 UGB Solution.

Solution:

Let r be a positive rational number. Define a_1=[r], so that

    \[0\leq r-a_1<1.\]

Assume that a_1,\ldots,a_m are defined in such a way that

    \[0\leq r-\sum^m_{k=1}\frac{a_k}{k!}<\frac{1}{m!}.\]

Define

    \[a_{m+1}=\left[(m+1)!\left(r-\sum^m_{k=1}\frac{a_k}{k!}\right)\right].\]

Since

    \[0\leq (m+1)! \left(r-\sum^m_{k=1}\frac{a_k}{k!}\right)<m+1,\]

we necessarily have a_{m+1}\in \{0,\ldots,m\}. So, there exists a sequence of integers (a_n:n\geq 1) such that

    \[a_1\geq 0, 0\leq a_n\leq n-1, \ n>1\]

and for all n\geq 1

    \[0\leq r-\sum^n_{k=1}\frac{a_k}{k!}<\frac{1}{n!}.\]

For large enough n we have n!r\in \mathbb{Z}, so

    \[n! \left(r-\sum^n_{k=1}\frac{a_k}{k!}\right)\in [0,1)\cap \mathbb{Z}.\]

The latter is only possible when

    \[r=\sum^n_{k=1}\frac{a_k}{k!}.\]

Now assume that there are two representations

    \[\sum^n_{k=1}\frac{a_k}{k!}=\sum^n_{k=1}\frac{b_k}{k!}\]

with integer coefficients a_k,b_k such that a_1,b_1\geq 0, a_k,b_k\in \{0,\ldots,k-1\} for k>1. Let j be the first index such that a_j\ne b_j. By symmetry we can assume that a_j<b_j. Then

    \[\frac{a_j}{j!}+\sum^n_{k=j+1}\frac{a_k}{k!}=\frac{b_j}{j!}+\sum^n_{k=j+1}\frac{b_k}{k!}.\]

Consider inequalities

    \[\frac{a_j}{j!}+\sum^n_{k=j+1}\frac{a_k}{k!}\leq \frac{a_j}{j!}+\sum^n_{k=j+1}\frac{k-1}{k!}=\]

    \[=\frac{a_j}{j!}+\sum^n_{k=j+1}\frac{1}{(k-1)!}-\sum^n_{k=j+1}\frac{1}{k!}=\]

    \[=\frac{a_j}{j!}+\frac{1}{j!}-\frac{1}{n!}<\frac{a_j+1}{j!}\leq \frac{b_j}{j!}+\sum^n_{k=j+1}\frac{b_k}{k!}.\]

Hence, the equality \sum^n_{k=1}\frac{a_k}{k!}=\sum^n_{k=1}\frac{b_k}{k!} is impossible.

Video Solution:

Problem 4.

Let g:(0, \infty) \rightarrow(0, \infty) be a differentiable function whose derivative is continuous, and such that g(g(x))=x for all x>0 . If g is not the identity function, prove that g must be strictly decreasing.

Topic: Calculus
Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 4 UGB Solution.

Solution:

We differentiate the relation g(g(x))=x:

    \[g'(g(x))g'(x)=1.\]

g'(x)\ne 0 for all x>0, hence g' preserves the sign on (0,+\infty). Assume that g'(x)>0 for all x>0. Then g(x) is strictly increasing. For some x we have g(x)\ne x (as g is not the identity function). If g(x)>x, then

    \[g(g(x))>g(x)>x,\]

but this contradicts g(g(x))=x. If g(x)<x, then

    \[g(g(x))<g(x)<x,\]

but this also contradicts g(g(x))=x. So, the assumption g'(x)>0 leads to a contradiction. Then g'(x)<0 for all x>0 and g is strictly decreasing.

Video Solution:

Problem 5.

Let a_{0}, a_{1}, \cdots, a_{19} \in \mathbb{R} and

    \[P(x)=x^{20}+\sum_{i=0}^{19} a_{i} x^{i}, \quad x \in \mathbb{R} .\]

If P(x)=P(-x) for all x \in \mathbb{R}, and

    \[P(k)=k^{2}, \text { for } k=0,1,2 \cdots, 9\]

then find

    \[\lim _{x \rightarrow 0} \frac{P(x)}{\sin ^{2} x}\]

Topic: Calculus
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 5 UGB Solution.

Solution:

The relation P(x)=P(-x) implies that

    \[x^{20}+\sum^{19}_{i=0}a_i x^i=x^{20}+\sum^{19}_{i=0}(-1)^ia_i x^i,\]

    \[\sum^{19}_{i=0} (1-(-1)^i)a_i x^i=0.\]

All coefficients a_i that correspond to odd indices i are equal to zero. So,

    \[P(x)=x^{20}+\sum^{9}_{i=0}a_{2i}x^{2i}.\]

Consider polynomial Q(x)=x^{10}+\sum^{9}_{i=0}a_{2i}x^{i}-x. By assumption,

    \[Q(k^2)=k^{20}+\sum^9_{i=0}a_{2i} k^{2i}-k^2=P(k)-k^2=0, 0\leq k\leq 9.\]

Observe that a_2-1 is the coefficient near x in Q(x). 10 roots of the polynomial Q are known:

    \[Q(x)=x (x-1)(x-2^2)\ldots (x-9^2).\]

In particular, a_0=0 and

    \[a_2-1=-\prod^9_{k=1} k^2, \ a_2=1-(9!)^2=-362879.\]

Then P(x)=-362879x^2+\sum^{19}_{i=3}a_i x^i+x^{20} and

    \[\lim_{x\to 0}\frac{P(x)}{\sin^2 x}=\lim_{x\to 0}\frac{x^2}{\sin^2 x}\lim_{x\to 0}\frac{P(x)}{x^2}=-362879.\]

Video Solution:

Problem 6.

If a given equilateral triangle \Delta of side length a lies in the union of five equilateral triangles of side length b, show that there exist four equilateral triangles of side length b whose union contains \Delta.

Topic: Geometry
Difficulty Level: Medium

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 6 UGB Solution.

Solution:

We show that necessarily b\geq \frac{a}{2}. Assume that b<\frac{a}{2}. Consider an equilateral triangle \Delta_b of side length b. The distance between any two points in \Delta_b is \leq b. Now consider an equilateral triangle \Delta_a of side length a. Consider the following configuration:

There are six points in the triangle with pairwise distances \geq \frac{a}{2}>b. Hence, if b<\frac{a}{2} then one needs at least six equilateral triangles of side length b to cover an equilateral triangle of side length a. So, b\geq \frac{a}{2} and the configuration above shows that 4 equilateral triangles of side length b are enough to cover an equilateral triangle of side length a.

Video Solution:

Problem 7.

Let a, b, c be three real numbers which are roots of a cubic polynomial, and satisfy a+b+c=6 and a b+b c+a c=9 . Suppose a<b<c . Show that

    \[0<a<1<b<3<c<4\]

Topic: Polynomials
Difficulty Level: Easy

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 7 UGB Solution.

Solution 1:

We have

    \[P(x)=(x-a)(x-b)(x-c)=x^3-x^2(a+b+c)+x(ab+ac+bc)-abc=x^3-6x^2+9x-abc.\]

The derivative of P(x):

    \[P'(x)=3x^2-12x+9=3(x-1)(x-3).\]

The polynomial P is increasing on (-\infty,1], decreasing on [1,3], increasing on [3,\infty). Then necessarily

    \[a<1<b<3<c.\]

Further,

    \[P(3)=27-54+27-abc=-abc<0,\]

i.e. abc>0 and a>0; and

    \[P(1)=4-abc>0, abc<4.\]

Finally,

    \[P(4)=64-24+36-abc=76-abc>72>0\]

and c<4.

Video Solution:

Problem 8.

A pond has been dug at the Indian Statistical Institute as an inverted truncated pyramid with a square base (see figure below). The depth of the pond is 6 \mathrm{~m}. The square at the bottom has side length 2 \mathrm{~m} and the top square has side length 8 \mathrm{~m} . Water is filled in at a rate of \frac{19}{3} cubic meters per hour. At what rate is the water level rising exactly 1 hour after the water started to fill the pond?

Topic: Calculus
Difficulty Level:

Indian Statistical Institute, ISI Subjective BStat & BMath 2021 Problem 8 UGB Solution.

Solution :

Denote by V(t) the volume of the water after t hours the water started to fill the pond. Then V(t)=\frac{19}{3}t (cubic meters). Let h(t) be the level of the water and a(t) be the side of the square at the top of the water after t hours the water started to fill the pond.

Extend the truncated pyramid to get the non-truncated pyramid and let x be the height of the added part. Then we have the following picture

Similarity of triangles implies relations

    \[\frac{x}{x+h(t)}=\frac{2}{a(t)}\]

    \[\frac{x+h(t)}{x+6}=\frac{a(t)}{8}.\]

Multiplying them we find

    \[\frac{x}{x+6}=\frac{1}{4}, \ x=2.\]

Hence,

    \[a(t)=h(t)+2.\]

Now,

    \[V(t)=\frac{h(t)}{3}(4+a(t)^2+2a(t))=\frac{h(t)}{3}(4+h(t)^2+4h(t)+4+2h(t)+4)=\]

    \[=\frac{h(t)}{3}(h(t)^2+6h(t)+12)=\frac{1}{3}((h(t)+2)^3-8).\]

Differentiate this:

    \[V'(t)=(h(t)+2)^2h'(t).\]

Then

    \[h'(1)=\frac{V'(1)}{(h(1)+2)^2}=\frac{V'(1)}{(3V(1)+8)^{2/3}}=\frac{\frac{19}{3}}{9}=\frac{19}{27} \mbox{ (meters per hour)}\]

Video Solution:

7 Comments

  • In two questions I messed up in the last line where we finally write the answer .
    Like instead of 19/27 or 19/3(3)^2 in the last question , is wrote 19/3(3) forgot the square , other wise all was fine until the last line …..
    And I’m polynomial question , I wrote 1-(9!)^2 instead of plus sign in last again , how much marks would that cost , can I get a 9/10 probably ,so email me

    • I feel sir made a mistake out there because the constant term of Q(x) is indeed -(9!)^2 as putting 0 in (x-1)(x-4)(x-9)……(x-81)….would give us (-1)(-4)(-9)….(-81) = -(9!)^2

      • Yes, there were some calculation errors. Thanks for notifying me. Corrected.

        • But sir the polynomial should be x^2(x^2 - 1^2)(x^2-2^2)...(x^2-9^2)+x^2. So the answer should be 1-(9!)^2

          But sir the polynomial should be x^2(x^2 – 1^2 )(x^2 – 2^2)… (x^2 – 9^2)+x^2.so the answer should be 1-(9!)^2.
          ( I have not given the exam this year, I read in class 12and will give isi entrance on 2022 though I have solved this in home and got this expression of the polynomial as result. Please inform me if it is wrong

  • sir, in the 4th question, since, g(g(x)) = x, g(x) is a function which is its own inverse. there are three such operations which have these property, which is, they cancel their own effect when acted an even number of times, which are as follows:
    (i) the identity function : doing nothing to the input.
    (ii) g(x) = -x : multiplication by -1 is cancelled by again multiplying it by -1.
    and (iii) taking the reciprocal, i.e., g(x) = 1/x or a/x.
    since, g(x) is defined from (o, infinity) to (o, infinity) and is not the identity function, it is the reciprocal i.e., g(x) = a/ x , for some positive, a.
    so its derivative is -a/x^2 which is always negative as x^2 and a are always positive. so, the function g(x) is strictly decreasing.

    is this correct, sir?

  • But sir the polynomial should be x^2(x^2 - 1^2)(x^2-2^2)...(x^2-9^2)+x^2. So the answer should be 1-(9!)^2

    But sir the polynomial should be x^2(x^2 – 1^2 )(x^2 – 2^2)… (x^2 – 9^2)+x^2.so the answer should be 1-(9!)^2.
    ( I have not given the exam this year, I read in class 12and will give isi entrance on 2022 though I have solved this in home and got this expression of the polynomial as result. Please inform me if it is wrong

  • Sir for problem 5, if we do not write the final answer(-362879) how much mark will be deduced?

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